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Derive an expression for the time period...

Derive an expression for the time period (T) of a simple pendulum which may depend upon the mass (m) of the bob, length (l) of the pendulum and acceleration due to gravity (g).

Text Solution

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Let `T= km^(a) I^(b) g^(c )`
where k is a dimensionless constant.
Writing the equation in dimensional form, we have `[M^(0) L^(0) T^(1) ] = [M]^(a) [L]^(b) [LT^(-2) ]^( c) = [M^(a) L^(b+c) T^(-2 c) ]`
Equating exponents of M, L and T on both sides,
we get `a=0, b+c = 0, -2c =1`.
Solving the eq., we get `a=0 , b=1//2 , c=-1//2`
Hence, `T= k sqrt((l)/(g))`, where `k` is constant.
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